Exam 1 Review

Use the following problems to prepare for the exam. There will be in-class review on Thursday, September 24. Your recitation this week will also be exam review.

Allowed Materials

You will be allowed to use a scientific calculator ( not a graphing calculator, not a calculator app on your phone). You may not share a calculator with another student; you must use your own calculator.

You may bring a standard 3 in x 5 in index card with prepared notes. You may use both sides of the notecard. You must put your full name in the top right corner of the card, and turn it in along with your exam.

Consider the sets A = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}, B = \{2, 4, 9, 10, 12, 14, 19\}, and C = \{9, 10, 11, 14, 16, 17, 20\}, which are all subsets of \Omega = \{1, 2, 3, \dotsc, 20\}.

Find A - (B \cap C).

\{1, 2, 3, 4, 5, 6, 7, 8\}

The set A - (B \cap C) consists of elements in the set A which are not in the overlap of B and C. The overlap is B \cap C = \{9, 10, 14\}, and of these elements, 9 and 10 are in A. So A - (B \cap C) = \{1, 2, 3, 4, 5, 6, 7, 8\}.

Find |A|, |B|, |C|, |A\cup B|, |A \cap B|, |B\cap C|, |A\cap C|, and |A\cup B\cup C|. Is it true that the size of the union of sets is equal to the sum of the sizes of the individual sets?

|A| = 10, |B| = 7, |C| = 7, |A \cup B| = 13, |A \cap B| = 4, |B\cap C| = 3, |A\cap C| = 2, |A\cup B\cup C| = 17. In particular, note that |A\cup B| = 13 \neq 10 + 7 = |A| + |B|, so it is not true in general that the size of the union of sets is the sum of the sizes of the individual sets.

|X| counts the number of elements in a finite set X. We can quickly count the elements in sets A, B, C to see that |A| = 10, |B| = 7, and |C| = 7.

The union of two sets includes all elements from either set, so: A \cup B \amp = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 19\} A \cup B \cup C \amp = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 14, 16, 17, 19, 20\} Counting the elements, we see |A \cup B| = 13 and |A \cup B \cup C| = 17.

The intersection of two sets is the overlap, consisting of elements which show up in both sets simultaneously, so: A \cap B \amp = \{2, 4, 9, 10\} B \cap C \amp = \{9, 10, 14\} A \cap C \amp = \{9, 10\} Counting the elements, we see |A\cap B| = 4, |B \cap C| = 3, and |A\cap C| = 2.

In particular, note that |A\cup B| = 13 \neq 10 + 7 = |A| + |B|, so it is not true in general that the size of the union of sets is the sum of the sizes of the individual sets.

Find A^c and (A\cup B)^c.

A^c = \{11, 12, 13, 14, 15, 16, 17, 18, 19, 20\}, (A\cup B)^c = \{11, 13, 15, 16, 17, 18, 20\}.

A^c consists of elements of \Omega which are not in A, thus A^c = \{11, 12, 13, 14, 15, 16, 17, 18, 19, 20\}. Similarly, (A\cup B)^c consists of elements that are not in A nor in B. Therefore, (A\cup B)^c = \{11, 13, 15, 16, 17, 18, 20\}.

Suppose we have a 6-sided die that's weighted to roll a 6 half of the time. We roll the die two times. List the set of all possible results. [Note: the result (2, 4)---rolling a 2 and then a 4---is different from the result (4, 2)---rolling a 4 and then a 2.]

\Omega = \{\amp (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), \amp (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), \amp (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), \amp (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), \amp (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), \amp (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)\}

Suppose we flip a coin two times. List the set of all possible results. What about flipping three times? Four times? If we flip the coin 10 times, how many possible results will there be?

For two flips: \Omega = \{ HH, HT, TH, TT \}.

For three flips: \Omega = \{ \amp HHH, HHT, HTH, THH, \amp HTT, THT, TTH, TTT \}

For four flips: \Omega = \{ \amp HHHH, HHHT, HHTH, HTHH, \amp THHH, HHTT, HTHT, HTTH, \amp THHT, THTH, TTHH, HTTT, \amp THTT, TTHT, TTTH, TTTT \}.

Each additional flip doubles the number of outcomes. So, with ten flips, we'll have |\Omega| = 2^{10} = 1024.

If we roll a 6-sided die ten times, how many possible results will there be?

|\Omega| = 6^{10}.

Each additional roll will multiply the number of outcomes by 6. So, with 10 rolls, we'll have |\Omega| = 6^{10}.

Consider the sample space \Omega = \{1, 2, 3, 4, 5, 6, 7, 8\} with probability distribution below. Calculate the probabilities of A = \{1, 3, 7, 8\}, B = \{2, 3, 6, 7\}, A\cup B, and A \cap B.

<tabular halign="center"> <row bottom="minor"> <cell><m>x</m></cell> <cell><m>\Pr(x)</m></cell> </row> <row> <cell>1</cell> <cell>0.1</cell> </row> <row> <cell>2</cell> <cell>0.05</cell> </row> <row> <cell>3</cell> <cell>0.2</cell> </row> <row> <cell>4</cell> <cell>0.15</cell> </row> <row> <cell>5</cell> <cell>0.15</cell> </row> <row> <cell>6</cell> <cell>0.1</cell> </row> <row> <cell>7</cell> <cell>0.05</cell> </row> <row> <cell>8</cell> <cell>0.1</cell> </row> <row> <cell>9</cell> <cell>0.1</cell> </row> </tabular> </table> </statement> <hint> Remember that, to calculate the probability of an event, you should add up the probabilities of each outcome in the event. </hint> <answer> <m>\Pr(A) = 0.45, \Pr(B) = 0.4, \Pr(A \cup B) = 0.6, \Pr(A \cap B) = 0.25.</m> </answer> <solution> <md> <mrow> \Pr(A) \amp = \Pr(1) + \Pr(3) + \Pr(7) + \Pr(8) </mrow> <mrow> \amp = 0.1 + 0.2 + 0.05 + 0.1 </mrow> <mrow> \amp = 0.45 </mrow> <mrow> \Pr(B) \amp = \Pr(2) + \Pr(3) + \Pr(6) + \Pr(7) </mrow> <mrow> \amp = 0.05 + 0.2 + 0.1 + 0.05 </mrow> <mrow> \amp = 0.4 </mrow> </md> The other events are <m>A\cup B = \{1, 2, 3, 6, 7, 8\}</m> and <m>A\cap B = \{3, 7\}</m>, so: <md> <mrow> \Pr(A \cup B) \amp = \Pr(1) + \Pr(2) + \Pr(3) + \Pr(6) + \Pr(7) + \Pr(8) </mrow> <mrow> \amp = 0.1 + 0.05 + 0.2 + 0.1 + 0.05 + 0.1 </mrow> <mrow> \amp = 0.6 </mrow> <mrow> \Pr(A \cap B) \amp = \Pr(3) + \Pr(7) </mrow> <mrow> \amp = 0.2 + 0.05 </mrow> <mrow> \amp = 0.25 </mrow> </md> </solution> </exercise> <exercise> <introduction> Suppose we flip a coin two times. Answer the questions below. What about three flips? What about four flips? </introduction> <task> <statement> Write all outcomes in the sample space <m>\Omega</m>. </statement> <answer> <m>\Omega = \{HH, HT, TH, TT\}.</m> </answer> </task> <task> <statement> Make a probability distribution table for <m>\Omega</m> assuming the coin is fair. </statement> <answer> <table> <title>Probability Distribution for Two Fair Coin Flips x \Pr(x) HH 0.25 HT 0.25 TH 0.25 TT 0.25

Make a probability distribution table assuming the coin comes up heads with probability 0.3.

Probability Distribution for Two Fair Coin Flips x \Pr(x) HH 0.09 HT 0.21 TH 0.21 TT 0.49

Suppose we roll a fair 6-sided die two times. Answer the questions below.

Write all outcomes in the sample space \Omega.

Make a probability distribution table for \Omega assuming the die is fair.

We'll avoid an overly large table and note that, since the die is fair, every outcome is equally likely. Therefore, \Pr(x) = \frac{1}{36} for every x\in \Omega.

Let A be the event that the second roll is higher than the first, and let B be the event that the first roll is even. Find \Pr(A), \Pr(B), and \Pr(A \cap B).

\Pr(A) = \frac{5}{12}, \Pr(B) = \frac{1}{2}, \Pr(A \cap B) = \frac{1}{6}.

A = \{ \amp (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), \amp (2, 3), (2, 4), (2, 5), (2, 6), \amp (3, 4), (3, 5), (3, 6), \amp (4, 5), (4, 6), \amp (5, 6)\} B = \{ \amp (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), \amp (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), \amp (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)\} A \cap B = \{ \amp (2, 3), (2, 4), (2, 5), (2, 6), \amp (4, 5), (4, 6)\} Therefore \Pr(A) = \frac{15}{36} = \frac{5}{12}, \Pr(B) = \frac{18}{36} = \frac{1}{2}, \Pr(A \cap B) = \frac{6}{36} = \frac{1}{6}.

Suppose a die has the values 1, 2, 3, 4, 5, 6 on the faces, but the die is not fair. Instead, the probabilities scale by the same amount as the face values. For example, a result of 4 is twice as likely as a result of 2, since 4 is twice as large as 2; a result of 6 is six times more likely than a result of 1; and so on. Write a probability distribution table for this die.

Probability Distribution for a Linearly Scaled Die x \Pr(x) 1 1/21 2 2/21 3 3/21 4 4/21 5 5/21 6 6/21

Let \Pr(1) = x. Then \Pr(2) = 2 \Pr(1) = 2x, and \Pr(3) = 3\Pr(1) = 3x, and so on. So the total probability in the space is: Pr(\Omega) \amp = \Pr(1) + \Pr(2) + \Pr(3) + \Pr(4) + \Pr(5) + \Pr(6) \amp = x + 2x + 3x + 4x + 5x + 6x \amp = 21x Since the total probability must add up to 1, we have 1 = 21x, so x = \frac{1}{21}, and we can calculate the rest of the probabilities from there.

Probability Distribution for a Linearly Scaled Die x \Pr(x) 1 1/21 2 2/21 3 3/21 4 4/21 5 5/21 6 6/21

Suppose a die has the values 1, 2, 3, 4, 5, 6 on the faces, but the die is not fair. Instead, each even value has an equal probability, each odd value has an equal probability, and the even values are each twice as likely as the odd values to appear on a roll. Write a probability distribution table for this die.

Probability Distribution for an Even-biased Die x \Pr(x) 1 1/9 2 2/9 3 1/9 4 2/9 5 1/9 6 2/9

Let \Pr(1) = x. Then \Pr(3) = x and \Pr(5) = x, since all odd rolls must have the same probability. The even rolls must have twice the probability, so \Pr(2) = \Pr(4) = \Pr(6) = 2x. Now: Pr(\Omega) \amp = \Pr(1) + \Pr(2) + \Pr(3) + \Pr(4) + \Pr(5) + \Pr(6) \amp = x + 2x + x + 2x + x + 2x \amp = 9x Since the total probability must add up to 1, we have 1 = 9x, so x = \frac{1}{9}, and we can calculate the rest of the probabilities from there.

Probability Distribution for an Even-biased Die x \Pr(x) 1 1/9 2 2/9 3 1/9 4 2/9 5 1/9 6 2/9

A toxin molecule inside a cell has a 0.3 probability of leaving the cell during a 1-minute period. For each value of n = 1, 2, 3, \dotsc, find the probability of the toxin molecule leaving the cell during the n th minute. What is the probability of the molecule leaving the cell during the first 3 minutes?

Write T for the minute that the toxin molecule leaves the cell. Then \Pr(T = n) = (0.7)^{n - 1} (0.3), and \Pr(T \leq 3) = 0.657.

Write T for the minute that the toxin molecule leaves the cell. We're told that the toxin molecule has probability 0.3 of leaving during each minute. Therefore, the probability of remaining during a particular minute is 1 - 0.3 = 0.7.

For the toxin molecule to leave the cell during minute n, it must remain for minutes 1, 2, \dotsc, n - 1, and then leave during minute n. The probability to remain each minute is 0.7, so we must multiply n - 1 copies of 0.7. Then the probability of leaving is 0.3, so we multiply by a factor of 0.3, yielding the formula: \Pr(T = n) = (0.7)^{n - 1}(0.3).

Then, the probaiblity of the molecule leaving within the first three minutes will be: \Pr(T \leq 3) \amp = \Pr(T = 1) + \Pr(T = 2) + \Pr(T = 3) \amp = 0.3 + (0.7)(0.3) + (0.7)^2(0.3) \amp = 0.657.

In each of the following scenarios with given events A and B, calculate \Pr(A), \Pr(B), \Pr(A\cap B), \Pr(A \mid B), and \Pr(B \mid A).

An experiment consists of rolling a fair die two times. Let A be the event that the sum is even, and let B be the event that the second roll is higher than the first.

A = \{ \amp (1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), \amp (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (4, 6), \amp (5, 1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6)\} B = \{ \amp (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), \amp (2, 3), (2, 4), (2, 5), (2, 6), \amp (3, 4), (3, 5), (3, 6), \amp (4, 5), (4, 6), \amp (5, 6)\} A \cap B = \{ \amp (1, 3), (1, 5), (2, 4), (2, 6), (3, 5), (4, 6)\} So \Pr(A) = \frac{18}{36} = \frac{1}{2}, \Pr(B) = \frac{15}{36} = \frac{5}{12}, and \Pr(A\cap B) = \frac{6}{36} = \frac{1}{6}. Finally: \Pr(A \mid B) \amp = \frac{\Pr(A\cap B)}{\Pr(B)} = \frac{6/36}{15/36} = \frac{6}{15} = \frac{2}{5} \Pr(B \mid A) \amp = \frac{\Pr(B\cap A)}{\Pr(A)} = \frac{6/36}{18/36} = \frac{6}{18} = \frac{1}{3}

An experiment consists of flipping a fair coin three times. Let A be the event that the first and second flips match. Let B be the event that there are at least two heads.

A \amp = \{ HHH, HHT, TTH, TTT \} B \amp = \{ HHH, HHT, HTH, THH \} A\cap B \amp = \{HHH, HHT\} So \Pr(A) = \frac{4}{8} = \frac{1}{2}, \Pr(B) = \frac{4}{8} = \frac{1}{2}, and \Pr(A\cap B) = \frac{2}{8} = \frac{1}{4}. Finally: \Pr(A \mid B) \amp = \frac{\Pr(A\cap B)}{\Pr(B)} = \frac{2/8}{4/8} = \frac{2}{4} = \frac{1}{2} \Pr(B \mid A) \amp = \frac{\Pr(B\cap A)}{\Pr(A)} = \frac{2/8}{4/8} = \frac{2}{4} = \frac{1}{2}

A diagnostic test is developed to detect a disease present in 3.2% of the population. For a patient who has the disease, the test will accurately give a positive result 65% of the time. When the patient does not have the disease, the test will accurately give a negative result 99.9% of the time.

For a patient who receives a positive test, what is the probability they have the disease?

\approx 0.96.

Let P be the event of testing positive and D the event of having the disease. Then the prevalence \Pr(D) is given as 3.2%, or 0.032. The sensitivity is \Pr(P\mid D) = 0.65, and the specificity is \Pr(P^c\mid D^c) = 0.999. So, according to Bayes' Theorem: \Pr(D\mid P) \amp = \frac{\Pr(P\mid D)\Pr(D)}{\Pr(P\mid D)\Pr(D) + (1 - \Pr(P^c\mid D^c))\Pr(D^c)} \amp = \frac{(0.65)(0.032)}{(0.65)(0.032) + (1 - 0.999)(1 - 0.032)} \amp \approx 0.96

For a patient who receives a negative test, what is the probability they do not have the disease?

\approx 0.99.

\Pr(D^c\mid P^c) \amp = \frac{\Pr(P^c\mid D^c)\Pr(D^c)}{\Pr(P^c\mid D^c)\Pr(D^c) + (1 - \Pr(P\mid D))\Pr(D)} \amp = \frac{(0.999)(1 - 0.032)}{(0.999)(1 - 0.032) + (1 - 0.65)(0.032)} \amp \approx 0.99

An experiment consists of rolling a fair die two times. Let A be the event that the sum is even, and let B be the event that the second roll is higher than the first. Are A and B independent?

A and B are not independent.

An experiment consists of flipping a fair coin three times. Let A be the event that the first and second flips match. Let B be the event that there are at least two heads. Are A and B independent?

A and B are independent.

Let A = \{1, 2, 3\} and B = \{3, 4, 5\} be events in the sample space \Omega = \{1, 2, 3, 4, 5, 6\}. Create a probability distribution for \Omega so that A, B are independent.

Example Distribution x \Pr(x) 1 0.1 2 0.2 3 0.2 4 0.1 5 0.1 6 0.3

Now \Pr(A) = 0.5, \Pr(B) = 0.4, and \Pr(A\cap B) = 0.2 = (0.5)(0.4) = \Pr(A)\Pr(B), so A and B are independent.

An experiment consists of flipping a biased coin 20 times. If the coin comes up heads with probability p = 0.3, find the probability of seeing 5 heads. Find the probability of seeing up to (and including) 3 heads.

Let S be the number of heads. Then S \sim \Bin(20, 0.3), so: \Pr(S = 5) \amp = {20 \choose 5} (0.3)^5 (0.7)^{20 - 5} \amp = \frac{20!}{(5!)(15!)} (0.3)^5 (0.7)^{15} \amp = \frac{20 \times 19 \times 18 \times 17 \times 16}{5 \times 4 \times 3 \times 2 \times 1} (0.3)^5 (0.7)^{15} \amp = (19 \times 3 \times 17 \times 16) (0.3)^5 (0.7)^{15} \amp \approx 0.179

An experiment consists of flipping a coin repeatedly until we first see heads.

If the coin comes up heads with probability 0.4, what is the probability we'll see our first heads within three flips? What about precisely on the third flip?

Let T be the number of flips until we see heads. Then T is geometric with parameter p = 0.4, so: \Pr(T = k) \amp = (1-0.4)^{k-1}(0.4) = 0.6^{k-1} \cdot 0.4

Which flip has the highest chance of being the first flip to come up heads?

\Pr(T = k) = 0.6^{k-1} \cdot 0.4, so every additional flip multiplies the probability by 0.6. Therefore, the highest value for \Pr(T = k) occurs when k = 1, in which case \Pr(T = 1) = 0.4.

A particular store has an average of 20 customers each hour. During a 4-hour afternoon shift, what is the probability of serving 80 customers.

Let N be the number of customers seen during the afternoon shift. Since the store averages 20 customers per hour, it will average 80 per 4-hours. So N \sim \Poiss(20, 4) will have distribution \Pr(N = k) = \frac{80^{k}}{k!} e^{-80}. Therefore: \Pr(N = 80) = \frac{80^{80}}{80!} e^{-80} \approx 0.045.

A continuous random variable X taking values in [1, 4] has p.d.f. f(x) = k(x - \sqrt{x}) for some constant k.

What is the value of k?

k = \frac{6}{17}.

Find \Pr(2 \leq X \leq 3).

\Pr(2 \leq X \leq 3) \approx 0.325

A continuous random variable X taking values in [1, 2] has p.d.f. \displaystyle{f(x) = \frac{1}{2}\left(\frac{1}{x^2} + x\right)}. Find the c.d.f. F(x). Use your c.d.f. to find \Pr\left(1 \leq X \leq \frac{3}{2}\right).

\frac{1}{2}\left(\frac{x^2}{2} - \frac{1}{x}\right) + \frac{1}{4}. \Pr\left(1 \leq X \leq \frac{3}{2}\right) \approx 0.479.

A continuous random variable X taking values in [2, 3] has c.d.f. F(x) = \frac{x^3}{3} - x^2 + 4. Find the p.d.f. f(x).

f(x) = x^2 - 2x.

Let T \sim \Exp(3). Find \Pr(1 \leq T \leq 3) and \Pr(T \geq 0.5).

\Pr(1 \leq T \leq 3) \approx 0.050. \Pr(T \geq 0.5) \approx 0.223.

Consider X, Y with the joint distribution table below. Are X, Y independent?

Joint distribution for <m>X, Y</m> X = 0 X = 1 Y = 0 0.2 0.3 Y = 1 0.4 0.1

No. For example, \Pr(X = 0, Y = 0) \neq \Pr(X = 0)\Pr(Y = 0).

Suppose X, Y have the distributions: \Pr(X = 0) \amp = 0.1 \amp \Pr(Y = 0) \amp = 0.4 \Pr(X = 1) \amp = 0.4 \amp \Pr(Y = 1) \amp = 0.6 \Pr(X = 2) \amp = 0.5 Assuming X, Y are independent, write a joint distribution table.

Joint Distribution X = 0 X = 1 X = 2 Y = 0 0.04 0.16 0.2 Y = 1 0.06 0.24 0.3

Consider a random variable X with probability distribution below. Find \E(X).

Suppose we flip a coin n = 100 times, and let N count the number of heads.

If the coin comes up heads on a flip with probability p = 0.4, what is \E(N)?

40.

What if n = 80 and p = 0.6?

48.

What if n = 200 and p = 0.5?

100.

If \E(X) = 3, \E(Y) = -2, and \E(Z) = 1, what is \E(4X + 5Y - Z + 3)?

A continuous random variable X taking values in [0, 1] has p.d.f. f(x) = 2x. What is \E(X)?

2/3.

A continuous random variable X taking values in [-1, 1] has p.d.f. f(x) = \frac{3x^2}{2}. What is \E(X)?

A continuous random variable X taking values in [1, 4] has p.d.f. f(x) = \frac{4}{3x^2}. What is \E(X)?

\frac{4}{3}\ln(4) \approx 1.85.

A continuous random variable X taking values in [1, 4] has p.d.f. f(x) = k(x - \sqrt{x}) for some constant k. In a previous problem ( ), you found the value of k. Now, find \E(X).

\frac{258}{85} \approx 3.04.

A continuous random variable X taking values in [1, 2] has p.d.f. \displaystyle{f(x) = \frac{1}{2}\left(\frac{1}{x^2} + x\right)}. Find \E(X).

\frac{1}{2}\left( \ln(2) + \frac{7}{3}\right) \approx 1.513.

Consider a random variable X with probability distribution below. Find \Var(X).

<tabular halign="center"> <row header="yes" bottom="minor"> <cell><m>x</m></cell> <cell><m>\Pr(X = x)</m></cell> </row> <row> <cell>1</cell> <cell>0.1</cell> </row> <row> <cell>2</cell> <cell>0.05</cell> </row> <row> <cell>3</cell> <cell>0.2</cell> </row> <row> <cell>4</cell> <cell>0.15</cell> </row> <row> <cell>5</cell> <cell>0.15</cell> </row> <row> <cell>6</cell> <cell>0.1</cell> </row> <row> <cell>7</cell> <cell>0.05</cell> </row> <row> <cell>8</cell> <cell>0.1</cell> </row> <row> <cell>9</cell> <cell>0.1</cell> </row> </tabular> </table> </statement> <answer> <m>5.76</m>. </answer> </exercise> <exercise> <statement> Let <m>X</m> be a random variable taking the values 1, 2, 3, 4, 5. Write a distribution table for <m>X</m>, then use your table to write a distribution for <m>X^2</m>. Then, find <m>\Var(X)</m>. </statement> </exercise> <exercise> <statement> Suppose we flip a coin <m>n = 100</m> times, and let <m>N</m> count the number of heads. If the coin comes up heads on a flip with probability <m>p = 0.3</m>, what is <m>\Var(N)</m>? What if <m>n = 80</m> and <m>p = 0.6</m>? What if <m>n = 200</m> and <m>p = 0.5</m>? </statement> <answer> When <m>n = 100</m> and <m>p = 0.3</m>, <m>\Var(N) = 21</m>. When <m>n = 80</m> and <m>p = 0.6</m>, <m>\Var(N) = 19.2</m>. When <m>n = 200</m> and <m>p = 0.5</m>, <m>\Var(N) = 50</m>. </answer> </exercise> <exercise> <statement> If <m>\E(X) = 3</m>, <m>\Var(X) = 2</m>, what is <m>\E(X^2)</m>? </statement> <answer> <m>11</m>. </answer> </exercise> <exercise> <statement> A continuous random variable <m>X</m> taking values in <m>[0, 1]</m> has p.d.f. <m>f(x) = 2x</m>. What is <m>\Var(X)</m>? </statement> <answer> <m>1/18</m>. </answer> </exercise> <exercise> <statement> A continuous random variable <m>X</m> taking values in <m>[-1, 1]</m> has p.d.f. <m>f(x) = \frac{3x^2}{2}</m>. What is <m>\Var(X)</m>? </statement> <answer> <m>0.6</m>. </answer> </exercise> <exercise> <statement> A continuous random variable <m>X</m> taking values in <m>[1, 4]</m> has p.d.f. <m>f(x) = \frac{4}{3x^2}</m>. What is <m>\Var(X)</m>? </statement> <answer> <m>4 - \left(\frac{4}{3}\ln(4)\right)^2 \approx 0.583</m>. </answer> </exercise> <exercise> <statement> A continuous random variable <m>X</m> taking values in <m>[1, 4]</m> has p.d.f. <m>f(x) = k(x - \sqrt{x})</m> for some constant <m>k</m>. In a previous problem ( <xref ref="exercise-continuous-RV-find-k"/> ), you found the value of <m>k</m>. Now, find <m>\Var(X)</m>. </statement> <answer> <m>\frac{2307}{238} - \left(\frac{258}{85}\right)^2 \approx 0.48</m>. </answer> </exercise> <exercise> <statement> A continuous random variable <m>X</m> taking values in <m>[1, 2]</m> has p.d.f. <m>\displaystyle{f(x) = \frac{1}{2}\left(\frac{1}{x^2} + x\right)}</m>. Find <m>\Var(X)</m>. </statement> <answer> <m>\frac{29}{24} - \frac{1}{2}\ln(2) \approx 0.862</m>. </answer> </exercise> <exercise> <statement> Suppose a coin has an unknown probability of coming up heads. We perform the experiment in five independent trials, during which it takes 4, 5, 4, 3, and 6 flips to see our first heads in each trial. What is the maximum likelihood estimation for the probability of the coin coming up heads on a flip? </statement> <answer> <m>5/22 \approx 0.227</m>. </answer> </exercise> <exercise> <statement> A particular store owner wants to approximate the average hourly rate at which customers come into the store. They observe 80 customers enter during a particular 4-hour shift. What is the maximum likelihood estimation for the hourly customer rate? </statement> <answer> <m>20</m>. </answer> </exercise> <exercise> <statement> A radioactive material emits particles at an unknown probabilistic rate <m>\lambda</m> particles per minute. We observe particles emitted at times 1.1, 1.7, 1.3, 2.2, 1.9, and 1.8 minutes. Write the likelihood function <m>\mathcal{L}(\lambda)</m> based on this data. What is the maximum likelihood estimation for <m>\lambda</m>? </statement> <answer> <m>\mathcal{L}(\lambda) = \left( \lambda e^{-1.1\lambda} \right) \left( \lambda e^{-1.7\lambda} \right) \left( \lambda e^{-1.3\lambda} \right) \left( \lambda e^{-2.2\lambda} \right) \left( \lambda e^{-1.9\lambda} \right) \left( \lambda e^{-1.8\lambda} \right)</m>. The MLE is <m>0.6</m>. </answer> </exercise> <exercise> <statement> Suppose a parameter <m>\theta</m> takes values in <m>[0, 1]</m> with likelihood function <m>\mathcal{L}(\theta) = \sqrt{\theta} - \theta^2</m>. Find the maximum likelihood estimation of <m>\theta</m>. </statement> <answer> <m>\left(\frac{1}{4}\right)^{2/3} \approx 0. 37</m>. </answer> </exercise> </exercises> </section>